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Operators on Inner-Product Spaces Self-Adjoint: T∗ = T. Normal: TT

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Operators on Inner-Product Spaces Self-Adjoint: T∗ = T. Normal: TT Operators on Inner-Product Spaces Self-Adjoint: T ∗ = T . Normal: TT ∗ = T ∗T . Positive: Self-adjoint and hT v; vi ≥ 0 for all v. Isometry: jjT vjj = jjvjj for all v. Let V be a finite-dimensional inner-product space. We wish to character- ize those linear operators T : V ! V that have an orthonormal basis of eigenvectors. Complex Spectral Theorem: When V is a complex vector space, V has an orthonormal basis of eigenvectors with respect to a linear operator T if and only if T is normal. Proof: If V has an orthonormal basis of eigenvectors with respect to a linear operator T then T has a diagonal matrix representation A, which implies m(T ∗) = A∗ with respect to the same basis. Since the matrix representations of T and T ∗ are diagonal, they commute, hence T commutes with T ∗. Conversely, suppose TT ∗ = T ∗T . Since V is a complex vector space, T has an upper-triangular matrix representation A with respect to an orthonormal basis (method: apply Gram-Schmidt to an upper-triangular basis). It will suffice to show that this matrix is diagonal. It will further suffice to show 2 2 that jjRijj = jjCijj for each i. Let fu1; : : : ; ung be the orthonormal basis. ∗ 2 2 We must show jjT eijj = jjT eijj for all i. However, for any v 2 V , jjT ∗vjj2 = hT ∗v; T ∗vi = hTT ∗v; vi = hT ∗T v; vi = = hT v; T vi = jjT vjj2: Real Spectral Theorem: When V is a real vector space, V has an or- thonormal basis of eigenvectors with respect to a linear operator T if and only if T is self-adjoint. Proof: If V has an orthonormal basis of eigenvectors with respect to a real operator T then T has a diagonal matrix representation A which satisfies AT = A. This implies that T is self-adjoint. Conversely, suppose that a real operator T : V ! V is self-adjoint. We will define a complex normal operator S : W ! W , use the Complex Spectral 1 Theorem to find an orthonormal basis of eigenvectors for W , and use this to find an orthonormal basis of eigenvectors for V . Let W = V × V with scalar multiplication defined by (a + bi)(v; w) = (av − bw; aw + bv) and inner product defined by h(v1; w1); (v2; w2)i = (hv1; v2i + hw1; w2i) + i(−hv1; w2i + hw1; v2i): Let S : W ! W be defined by S(v; w) = (T v; T w): S is self-adjoint, hence normal, hence V × V has an orthonormal basis of eigenvectors f(v1; w1);:::; (vn; wn)g. The equation S(v; w) = (a + bi)(v; w) implies T (v) = av − bw and T (w) = aw + bv. Making these substitutions into hT v; wi = hv; T wi and rearranging yields b(hv; vi + hw; wi) = 0. This forces b = 0. Hence for each (vi; wi) there is a real number ai such that T vi = aivi and T wi = aiwi. Since f(v1; w1);:::; (vn; wn)g is a basis for V ×V over C, every (x; 0) is in the span of f(v1; w1);:::; (vn; wn)g over C, which implies that each x 2 V is in the span of fv1; w1; : : : ; vn; wng over R. Hence V has a basis which is a subset of fv1; w1; : : : ; vn; wng, hence V has a basis of eigenvectors of T . Now let Ba be the set of those basis eigenvectors corresponding to eigen- value a. Applying Gram-Schmidt to these produces an orthonormal set of eigenvectors Oa corresponding to eigenvalue a. If u and v are eigenvectors corresponding to distinct eigenvalues of T then they are orthogonal to each other: suppose T u = au and T v = bv where a 6= b. Then ahu; vi = hau; vi = hT u; vi = hu; T vi = hu; bvi = bhu; vi: This forces hu; vi = 0. The union of the Oa bases forms an orthonormal eigenvector basis of V . Remark: Taken together, the two spectral theorems yield an algorithm for finding an orthonormal basis of eigenvectors for an inner product space V given a linear operator T : V ! V which is normal (V complex) or self- adjoint (V real): First, find all the eigenvalues of T . Second, compute each 2 eigenspace. Since V has in principle an orthonormal basis of eigenvectors, V is the sum of its eigenspaces. Since eigenvectors corresponding to distinct eigenvalues are linearly independent, the sum is direct. Hence the union of the eigenspace bases forms a basis of eigenvectors for V . When V is complex and T is normal we know that applying Gram-Schmidt to a basis of eigenvectors produces an orthonormal basis of eigenvectors since the original basis produces an upper-triangular matrix representation. When V is real and T is self-adjoint we know that eigenvectors corresponding to distinct eigenvalues are orthogonal, so applying Gram-Schmidt to a eigenvector basis produces an orthonormal eigenvector basis. To summarize, in either case we find a basis for each eigenspace, then apply Gram-Schmidt to the union of these bases. Characterization of Real Normal Linear Operators: Let T : V ! V be a normal operator on a real inner product space V . We will show that T has a block-diagonal matrix representation with respect to an orthonormal basis, where each of the blocks has dimension ≤ 2 and each of the blocks a −b of dimension 2 has the form . We can use an induction argument: b a for dimension 1 this is trivial. More generally, find an invariant subspace U of dimension ≤ 2, find an orthonormal basis for it, and expand to an orthonormal basis for V . The matrix representation of T restricted to U has the desired form, using the algebra steps which appear in the proof of the Complex Spectral Theorem. The same algebra steps can be used to show A 0 that the matrix representation of T with respect to this basis is 1 . 0 B So we have the foundation for an induction argument: we have decomposed V into U L U ? where T maps U into U and U ? into U ? and the matrix representation of T restricted to these invariant subspaces are A1 and B and U has dimension ≤ 2. The induction hypothesis enables us to find an orthonormal basis for U ? of the desired form, and adding these vectors to the orthonormal basis for U produces the desired orthonormal basis for V . Conversely, if T has such a matrix representation with respect to an orthonormal basis then it is normal. Note also that T is normal but not self- a −b adjoint if and only if T has at least one 2 × 2 block of the form with b a b 6= 0. By replacing one basis vector with its additive inverse if necessary we can assume b > 0. 3 a11 a12 (Note: if T is represented by with respect to the basis fu1; u2g a21 a22 a (β/α)a then it is represented by the matrix 11 12 with respect to (α/β)a21 a22 the basis fαu1; βu2g.) Characterization of Positive Operators: An operator T : V ! V on a complex or real finite-dimensional inner product space is said to be positive if it is self-adjoint and satisfies hT v; vi ≥ 0 for each v 2 V . We know that the self-adjoint operators are precisely those that have a diagonal matrix representation with respect to some orthonormal basis of eigenvectors, where the diagonal entries rii are real numbers. The equation X X X 2 hT ( αiei); αieii = riijαij ≥ 0 i i i holds for all choices of (α1; : : : ; αn), which happens if and only if the diag- onal entries are non-negative real numbers. So the positive operators are precisely those that have a diagonal matrix representation with respect to some orthonormal basis of eigenvectors, in which the diagonal entries (i.e. the eigenvalues) are all non-negative. Theorem 7.27 is a list of equivalent conditions for a positive operator T . Given this characterization, each positive operator has a positive square root: just use the same orthonormal basis of eigenvectors, but let the matrix rep- resentation be that in which the the diagonal entries are replaced by their non-negative square roots. This square root is unique: let T be a positive 2 operator and let S be a positive operatorp that satisfies S = T . Computing S on the eigenvectorp basis yields Sv = λv where T v = λv. This determines S. Notation: S = T . For any linear operator T : V ! V on an inner product space V , the operator ∗ ∗ 2 T T is always positive:p hT T v; vi =phT v; T vi = jjT vjj ≥ 0. Hence one can always construct T ∗T . The map T ∗T has some interesting properties: p p 1. For all u; v 2 V , h T ∗T u; T ∗T vi = hT u; T vi. Proof: p p h T ∗T u; T ∗T vi = hT ∗T u; vi = hT u; T vi: p 2. null( T ∗T ) = null(T ). Proof: First note that null(T ) ⊆ null(T ∗T ). On the other hand, if u 2 null(T ∗T ) then T ∗T u = 0, hence hT ∗T u; ui = 0, 4 therefore hT u; T ui = 0, therefore jjT ujj2 = 0, therefore T u = 0, there- ∗ fore u 2 null(T ). Hence null(T ) = null(T T ). Using the diagonalp matrix ∗ ∗ ∗ representationp of T T we can show that null(T T ) = null( T T ). Hence null(T ) = null( T ∗T ). p 3.
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